exponential distribution involving customers 
Customers arrive at a certain shop according to a poisson process at
  the rate of $20$ customers per hour. what is the probability that the
  shop keeper will have to wait more than $5$ minutes (after opening)
  for the arrival of the first customers?

So i am trying to this problem by exponential, however im having some trouble with figuring out the pdf.
I was told $F(x)=20e^{-20x}$ if $x>0$, however the pdf on my paper says: $\frac{1}{\theta}e^{-\frac{x}{\theta}}$ so shouldnt it be: $\frac{1}{20}e^{-\frac{x}{20}}$.
I am just confused. Any explanations?
 A: Well the Poisson process and exponential variables are related. Imagine the exponential as a general expression for amount of time for the next occurrence of event A, and imagine Poisson as the amount of occurrences of event A in a general time period. 
For example, 20 customers come in per hour. This means one customer for every 3 minutes. This is known as the mean= ø
Thus your expression should be $\cfrac{e^{\frac{x}{3}}}{3}$.
Choose integration bounds [5, infinity), or use the complement of the integral for the above expression with bounds [0,5]
A: In a Poisson process with rate $\lambda$ per unit of time, the interarrival times between events is exponentially distributed with mean $1/\lambda$.  Therefore, the time to the first arrival $T_1$ is exponential with probability density $$f_{T_1} (t) = \lambda e^{-\lambda t}, \quad t > 0.$$  The cumulative distribution function is $$\Pr[T_1 \le t] = 1 - e^{-\lambda t}, \quad t > 0.$$  What you desire is $$\Pr[T_1 > 1/12],$$ because five minutes is equivalent to one twelfth of an hour.
A: Let the discrete random variable $X=$ The number of minutes the shopkeeper will have to wait for the first customers to arrive.
$X\sim Po(\lambda)$ and working in minutes $\implies 3 \space$  customers arrive per minute, such that $\lambda=3$ so $$X\sim Po\left(3\right)$$ $$P\left(X \gt 5\right)=1-P\left(X \le 5\right) \approx \color{purple}{0.084}$$ 
